# Total electric charge on the earth,

1. Aug 23, 2010

### vorcil

1. The problem statement, all variables and given/known data
The earth can be regarded as a sphere of radius R,
the volume density of charge distributed as $$\rho = \frac{\rho_s r}{R}$$
where the density is 0 at thcce centre and rises linearly with radius until it reaches ps at the earth's surface

i) prove that the total charge on the earth is $$\pi \rho_s R^3$$

ii) use gauss's law to find an expression for E at the point r outside the earth (r>R)

iii) use gauss's law to find an expression for E within the earths interior (r<R)

iv) write expressions for the x-, y- and z- components of E at a point outside the earth and calculate $$\bf{\nabla .E } 2. Relevant equations Qenclosed = [tex] \int_v \rho d\tau.$$ (for a volume)

$$\oint_S \bf{E} . d\bf{a} = \frac{1}{\epsilon} Qenclosed$$

3. The attempt at a solution

$$\int_v \rho \dtau$$

substituting in the given rho,

$$\int \rho_s \frac{r}{R} \dtau$$

the outer radius r is = R so it just becomes R/R =1,

$$\rho_s *1 \int \dtau$$

this is where I integrate the dtau and it should give me the volume of a sphere

$$\rho_s \frac{4}{3}\pi r^3$$

this sort of looks like the answer i'm looking for
HOWEVER I DONT KNOW HOW TO GET RID OF THE 4/3 constant!!!
what have I done wrong?
i've tried heaps of other methods, but I'm always getting some type of constant out the front

2. Aug 23, 2010

### phyzguy

The density is proportional to r, as you wrote, but you can't pull this factor of r outside the integral. You should write:
$$\int_0^R\rho(r)d\tau = \int_0^R\rho_s\frac{r}{R}4\pi r^2 dr = 4\pi\frac{\rho_s}{R}\int_0^Rr^3 dr = 4\pi \frac{\rho_s}{R} \frac{R^4}{4} = \pi\rho_sR^3$$

Does this make sense?

3. Aug 23, 2010

### vorcil

Yes thanks, I actually ended up doing it like this myself last night, I just didn't update it here on PF